Puzzle. The professor is watching across a field how the son of the professor’s father is fighting with the father of the professor’s son. How is this possible?

This puzzle is tricky only because of gender-bias. Most people assume that the professor is male and miss the obvious intended solution, in which a female professor is watching her brother fighting with her husband.

I just gave this problem on a test. Here are other answers that I received.

The professor is gay and is watching his brother fighting with his husband.

The professor is watching his brother fighting with the father of the professor’s step-son.

The father of the professor’s son is himself. So he is watching a video of himself fighting with his brother.

Years ago people couldn’t figure out this puzzle at all. So there has been progress. I was glad that my students suggested so many ideas that work. Nonetheless, many of them revealed their gender-bias by initially assuming that the professor is a man.

Puzzle. There are five houses of different colors next to each other equally spaced on the same road. In each house lives a man of a different profession.

The blue house is adjacent to the mathematician’s and con-man’s houses.

The first house on the left is green.

The nurse lives immediately to the right of the mathematician.

The teacher lives halfway between the plumber’s house and the yellow house.

The nurse’s house is immediately to the right of the red house.

Who lives in the white house?

Correction Nov 11, 2017. Replaced “the same distance from” with “halfway between” to eliminate the possibility of the plumber living in the yellow house. Thank you to my readers for catching this mistake and to Smylers for suggesting a correction.

Problem. Invent a connected shape made out of squares on the square grid that cannot be cut into dominoes (rectangles with sides 1 and 2), but if you add a domino to the shape then you can cut the new bigger shape.

This problem reminds me of another famous and beautiful domino-covering problem.

Problem. Two opposite corner squares are cut out from the 8 by 8 square board. Can you cover the remaining shape with dominoes?

The solution to the second problem is to color the shape as a chess board and check that the number of black and white squares is not the same.

What is interesting about the first problem is that it passes the color test. It made me wonder: Is there a way to characterize the shapes on a square grid that pass the color test, but still can’t be covered in dominoes?

I already wrote about two puzzles that Derek Kisman made for the 2013 MIT Mystery Hunt. The first puzzle is now called the Fractal Word Search. It is available on the Hunt website under its name In the Details. I posted one essay about the puzzle and another one describing its solution. The second puzzle, 50/50, is considered one of the most difficult hunt puzzles ever. Unfortunately, the puzzle is not available, but my description of it is.

Today let’s look at the third puzzle Derek made for the 2013 Hunt, building on an idea from Tom Yue. This is a non-mathematical crossword puzzle. Derek tends to write multi-layered puzzles: You think you’ve got the answer, but the answer you’ve got is actually a hint for the next step.

Often multi-layered puzzles get solvers frustrated, but the previous paragraph is a hint in itself. If you expect the difficulty, you might appreciate the fantastic beauty of this puzzle.

Problem. A coin collector has 100 coins that look identical. He knows that 30 of the coins are genuine and 70 fake. He also knows that all the genuine coins weigh the same and all the fake coins have different weights, and every fake coin is heavier than a genuine coin. He doesn’t know the exact weights though. He has a balance scale without weights that he can use to compare the weights of two groups with the same number of coins. What is the smallest number of weighings the collector needs to guarantee finding at least one genuine coin?

Now it’s solution time. First we show that we can do this in 70 weighings. The strategy is to compare one coin against one coin. If the scale balances, we are lucky and can stop, because that means we have found two real coins. If the scale is unbalanced, the heavier coin is definitely fake and we can remove it from consideration. In the worst case, we will do 70 unbalanced weighings that allow us to remove all the fake coins, and we will find all the real coins.

The more difficult part is to show that 69 weighings do not guarantee finding the real coin. We do it by contradiction. Suppose the weights are such that the real coin weighs 1 gram and the i-th fake coin weighs 100i grams. That means whatever coins we put on the scale, the heaviest pan is the pan that has the fake coin with the largest index among the fake coins on the scale.

Suppose there is a strategy to find a real coin in 69 weighings. Given this strategy, we produce an example designed for this strategy, so that the weighings are consistent, but the collector cannot find a real coin.

For the first weighing we assign the heaviest weight, 10070 to one of the coins on the scale and claim that the pan with this coin is heavier. We continue recursively. If a weighing has the coins with assigned weights, we pick the heaviest coin on the pans and claim that the corresponding pan is heavier. If there are no coins with assigned weights on pans, we pick any coin on the pans, assigned the largest available weight to it and claim that the corresponding pan is heavier.

After 69 weighings, not more than 69 coins have assigned weights, while all the weighings are consistent. The rest of the coins can have any of the leftover weights. For example, any of the rest of the coins can weigh 100 grams. That means that there is no coin that is guaranteed to be real.

This problem is really very difficult. The competition organizers offered an extra point for every diagonal on top of 16. The official solution has 24 diagonals, but no proof that it’s the longest. I’m not sure anyone knows if it is the longest.

Here is another problem:

Problem. Alice and Bob are playing a game. They start with two numbers: 2014 and 2015. In one move a player can do one of two things:

subtract one of the numbers by one of the non-zero digits in any of the two numbers or,

divide one number by two if the number is even.

The winner is the person who is the first to get a one-digit number. Assuming that Alice starts, who wins?

I like Odd-One-Out puzzles that are ambiguous. That is why I bought the book Which One Doesn’t Belong? Look at the cover: which is the odd one out? The book doesn’t include answers, but it has nine more examples in each of which there are several possible odd-one-outs.

Problem. A store sells letter magnets. The same letters cost the same and different letters might not cost the same. The word ONE costs 1 dollar, the word TWO costs 2 dollars, and the word ELEVEN costs 11 dollars. What is the cost of TWELVE?

The beautiful Pascal triangle has been around for many years. Can you say something new about it?

Of course you can. Mathematicians always find new way to look at things. In 2012 RSI student, Kevin Garbe, did some new and cool research related to the triangle. Consider Pascal’s triangle modulo 2, see picture which was copied from a stackexchange discussion.

A consecutive block of m digits in one row of the triangle modulo 2 is called an m-block. If you search the triangle you will find that all possible binary strings of length 2 are m-blocks. Will this trend continue? Yes, you can find any possible string of length 3, but it stops there. The blocks you can find are called accessible blocks. So, which blocks of length 4 are not accessible?

There are only two strings that are not accessible: 1101 and 1011. It is not surprising that they are reflections of each other. Pascal’s triangle respects mirror symmetry and the answer should be symmetric with respect to reflection.

You can’t find these blocks on the picture, but how do we prove that they are not accessible, that is, that you can’t ever find them? The following amazing property of the triangle can help. We call a row odd/even, if it corresponds to binomial coefficients of n choose something, where n is an odd/even number. Every odd row has every digit doubled. Moreover, if we take odd rows and replace every double digit with its single self we get back Pascal’s triangle. Obviously the two strings 1101 and 1011 can’t be parts of odd rows.

What about even rows? The even rows have a similar property: every even-indexed digit is a zero. If you remove these zeros you get back Pascal’s triangle. The two strings 1101 and 1011 can’t be part of even rows. Therefore, they are not accessible.

The next question is to count the number of inaccessible blocks of a given length: a(n). This and much more was done by Kevin Garbe for his RSI 2012 project. (I was the head mentor of the math projects.) His paper is published on the arxiv. The answer to the question can be found by constructing recurrence relations for odd/even rows. It can be shown that a(2r) = 3a(r) + a(r+1) − 6 and a(2r+1) = 3a(r) + 2a(r+1) − 6. As a result the number of inaccessible blocks of length n is n2 − n + 2. I wonder if there exists a direct proof of this formula without considering odd and even rows separately.

This RSI result was so pretty that it became a question at our entrance PRIMES test for the year 2013. In the test we changed the word accessible to admissible, so that it would be more difficult for applicants to find the research. Besides, Garbe’s paper wasn’t arxived yet.

The pretty picture above is from the stackexchange, where one of our PRIMES applicants tried to solicit help in solving the test question. What a shame.